ASSIGNMENT SHEET #4 PART I APQ ANSWERS
5
a. (Recall: combustion means adding oxygen gas to)
C5H12 + 8 O2 → 5 CO2 + 6 H2O
b. 2.50 g C5H12 ÷ 72.15 g/mole C5H12 = 0.035 mole C5H12
0.035 mole C5H12 x
5moleCO2
= 0.173 moles CO2
1moleC5 H12
Using PV = nRT, V =
€
nRT (0.173mole)(0.0821)(25 + 273K)
=
= 4.10 L
785mmHg
P
760mmHg /atm
c. 5.00 g C5H12 ÷72.15 g/mole C5H12 = 0.070 mole C5H12
€
€
243kJ
?kJ
Then use a ratio:
=
? = 3,471 kJ/mole
0.07mole 1mole
∴ ΔH
= - 3471 kJ/mol
d. (Recall: effusion is diffusion through a porous partition; use
Graham’s law)
€
rateunknown gas = 2 (ratepentane)
€
€
rateunknowngas
=
rate pen tan e
MM pen tan e
therefore, 2 =
MM unknowngas
72.15g
X = 18 g/mole
Xg
(Note: the unknown gas could be water!)
€
€
26 €
a. PV =
massgas
RT
MM gas
Pgas = Patm – Pwater (water vapor pressure, WVP)
b. Mass of gas container before allowing some gas to leave as well as the mass afterwards.
Read a thermometer and a barometer. Find the WVP at the given temperature. The volume of gas.
€
c. Equalizing the water levels inside and outside enables you to determine the pressure of the gas inside
of the gas collection tube. At that point, the pressures inside and outside are the same.
d.
€
| 64g − 58g |
x 100 = percent error
58g
e. If pressure is not corrected for WVP, the P value will be too large. Solving for molar mass, MM,
mass
(mass)(R)(T)
using the equation PV =
RT, you get MM =
. The larger value for P will give you
MM
PV
a smaller value for MM.
€
€
46
a. Each of the balloons contains the same number of moles of gas (according to Avogadro’s law).
Therefore, the balloon with the highest mass of gas contains the gas with the highest molar mass,
which is carbon dioxide (44 g/mole).
b. Since all of the balloons are at the same temperature, and since temperature is a measure of average
kinetic energy, the gases in each of the balloons have the same average kinetic energies.
c. Carbon dioxide gas would deviate the most from ideal gas behavior. It has the greatest relative
volume, so would occupy the most space. All of the gases are nonpolar compounds, which means they
only exhibit dispersion forces as IMFs. The strength of dispersion forces is dependent on the
size/volume of the compound; hence CO2’s largest volume also gives it the strongest dispersion forces.
1
d. Since average KE = mass x velocity2, and since all of the gases have the same average KE (they
2
are all at the same temperature), the gas with the lightest molar mass will have the greatest velocity,
and will thus escape the fastest through the pores of the balloon. Therefore, the He balloon will shrink
the most.
€
51
a. (Recall: combustion is the addition of oxygen gas)
C3H8 + 5 O2 → 3 CO2 + 4 H2O
b. 10.0 g C3H8 ÷ 44 g/mole C3H8 = 0.227 mole C3H8
0.227 mole C3H8 x
5moleO2
= 1.14 mole O2 needed to completely combust
1moleC3 H 8
Using PV = nRT, V =
€
(1.14mole)(0.0821)(30 + 273K)
= 28.27 L of O2 needed
1atm
100Lair
28.27 L O2 x
= 134.6 L air needed
21LO2
€
52
€ PV =
a. Using
mass
(MM)(P)(V )
RT, mass C02 produced =
MM
RT
750mmHg
75mL
)(
)
760mmHg /atm 1000mL /L
= 0.1354 g
€
(0.0821)(20 + 273K)
(44g /mole)(
Mass€CO2 =
€
68
a. Phydrogen gas = Patm – Pwater = 745 mmHg – 23.8 mmHg = 721.2 mmHg ÷ 760 mmHg/atm = 0.949 atm
90mL
(0.949atm)(
)
PhydrogenV
1000mL /L = 0.00349 moles H
Using PV = nRT, nhydrogen =
=
2
RT
(0.0821)(25 + 273K)
745mmHg
)(0.09L)
P
V
760mmHg
/atm
atm
€ ntotal gas = €
b. Using PV = nRT,
=
= 0.003606 moles gas
RT
(0.0821)(298K)
(
moles H2O = moles of total gas – moles of H2 gas = 0.003606 mole – 0.00349 mole = 0.000116 mole
€
0.000116 mole H2O x 6.02 x€1023 molecules/mole = 6.98 x 1019 water vapor molecules
c. Using Graham’s Law:
ratehydrogen
=
ratewater
MM water
=
MM hydrogen
18g /mole
=3
2g /mole
Therefore, the hydrogen molecules are moving 3 times as fast as the water molecules.
€
€
d. Water deviates more from€
ideal gas behavior than hydrogen because H2O has stronger IMFs due to
its polar nature and it has a larger volume due to its larger molar mass.
401
a. Gas behavior deviates from ideality at low temperatures and/or under high pressures. At low temps,
the gas particles move more slowly, allowing intermolecular forces (IMFs) to become significant
factors in particle motion. Under high pressures, the volumes of the individual particles become
significant – gases can’t be compressed into nothing. Under extremes of both situations, the gases tend
to liquefy or deposit (solidify).
b. Sulfur dioxide will deviate the most due to its polar nature – allowing for stronger IMFs (dipoledipole forces) to come into play – and due to its relatively large volume and terminal lone pairs.
c. Under low pressure, the space between the gas particles in a gas sample is so large that the volumes
of the individual particles are of no consequence. At high temperatures, the speed of the particles is so
great that no IMFs can possibly influence the motion of the individual particles.
402
The a value is largely determined by the gas particle’s polarity and the b value is determined by the gas
particle’s volume. H2S has the higher a value and b value due to its relative polarity (it exerts dipole
dipole forces, which are stronger IMFs than hydrogen’s dispersion forces) and volume, respectively,
compared to H2.
403
PV =
€
mass
3.53g
mmHg ⋅ L
RT = (750mmHg)(1L) =
(62.4
)(27 + 273)
MM
MM
mole ⋅ K
That’s All For Now, folks!
MM = 88.1 g/mol
ASSIGNMENT SHEET #4 PART II APQ ANSWERS
2d
The boiling point of chlorine is lower than that of bromine because of chlorine’s relatively
smaller mass and size. While both molecules are non-polar and have only dispersion forces
as IMFs, the larger volume bromine molecule dispersion forces are of greater strength than
those of the chlorine molecules. Hence, bromine molecules require more energy to overcome
the IMFs.
16
a. The polar NH3 molecules have hydrogen bonding as their principle IMF, which is an
appreciably stronger IMF than the dispersion forces of the non-polar CH4 molecules.
b. Both ethane and hexane are non-polar molecules, which means that they each have
dispersion forces as their IMFs. Dispersion forces are greater in molecules with greater
volume, so they are stronger in hexane than in ethane. Hence, hexane is held together in the
liquid state at room temperature, whereas ethane can overcome the IMFs at room
temperature and become gaseous.
c. Silicon can form network covalent bonds, which are much stronger than the dispersion
forces that hold chlorine molecules together.
d. The charge densities of the ions in MgO (+2 and –2 ions) are much stronger than the charge
densities of the ions in NaF (+1 and –1 ions). This creates stronger electrostatic forces
between the ions in MgO, which results in a higher melting point.
38a
Water boils when its vapor pressure equals atmospheric pressure. The atmospheric pressure
is higher at sea level than at high altitude, so the water must get hotter at sea level in order to
boil. Therefore, at high altitude, the water boils at a lower, temperature, requiring a longer
cooking time.
50a
HF molecules can form hydrogen bonds with one another; HCl molecules are held together
with dipole-dipole forces. The H-bonds are stronger than the dipole-dipole forces, so the HF
is a liquid whereas the HCl is a gas at the same temperature.
60
a. Point V represents the triple point. At the triple point of any substance, all three phases of
matter – solid, liquid, gas – exist in equilibrium with one another.
b. Every point on the curve between V and W represents an equilibrium condition between
the liquid and gaseous states; i.e., evaporation and condensation are occurring at the same
rate.
c. The material goes from a solid (X) to the sublimation point (Y), where the material
sublimates and deposits at the same rate, then becomes a gas (Z).
d. The solid will sink in the liquid. From the phase diagram, at a constant temperature, as
you increase the external pressure on the surface of the material, which makes the material
more dense, the material remains in the solid state or changes from a liquid to a solid. This
indicates that the solid is more dense than the liquid. The slope of the solid-liquid line is
positive, which means that as you increase the external pressure, you raise the freezing point
(temperature); i.e., making it “easier” to freeze the material – you don’t have to remove as
much energy to allow the IMFs to solidify the material. This is only true for substances
whose solid states are denser than their liquid states.
65bc
b. Being a more polar molecule, H2O molecules form stronger hydrogen bonds than the less
polar NH3 molecules. The two lone pairs on each of the oxygen atoms are more negative
than the single lone pair on each of the nitrogen atoms. Therefore, the hydrogen atoms are
more strongly attracted to the lone pairs in water than those in ammonia.
Additionally, each liquid water molecule can form four H-bonds per molecule on average,
whereas ammonia forms only two H-bonds per molecule on average.
c. In each material, the carbon atoms are network covalently bound. However, in diamond,
the network covalent bonding is 3-dimensional, whereas in graphite, the network covalent
bonding is 2-dimensional. Weak dispersion forces loosely hold the 2-dimensional sheets of
graphite atoms together. As such, the sheets shear from one another easily, and the sheets act
as lubricants. Diamond does not shear apart, and hence cannot be used as a lubricant.
67b
As the air pressure decreases in the bell jar, the boiling point will eventually be reached at the
lower room temperature. You will see the water boil. Over time, the beaker will be
completely empty as the pump removes the water vapor.
404
a. The metallic bonding of potassium allows for current conduction. The “sea of valence
electrons” is quite mobile and delocalized in terms of nuclear control. In the ionic bonds
found in solid potassium nitrate, the electrons are very localized – they are tightly held by
each ion. As such, there are no mobile electrons. Hence, the KNO3 is not a conductor of
current.
b. The non-polar carbon tetrachloride molecule is smaller in volume than the non-polar
carbon tetrabromide molecule. CCl4’s dispersion forces are therefore weaker than CBr4’s,
which explains its lower relative boiling point.
c. (You saw this Q in 50 a; the AP folks like this question!) HF has a stronger dipole moment
than HCl due to the greater difference in electronegativity in HF relative to that in HCl. HF
molecules can form hydrogen bonds that are stronger than the dipole-dipole forces found in
HCl. These stronger IMFs hold HF in the liquid state and give HF a higher boiling point.
d. Butane is more non-polar than chloroethane. Therefore, butane has only relatively weak
dispersion forces keeping it together whereas chloroethane and relatively strong dipoledipole forces holding it together.